This is material science question 3. Stress – Strain BehaviorConsider the brass alloy for which the stress-strain behavior is shown in figure Formulabelow. A cylinder specimen of this material 10 mm in diameter and 120 mm long is Tensile stress:pulled in tension with force of 15 kN. The Poisson's ratio is 0.35.Compute :F.OavgAShear stressa) The Tension stress of 15 kN force.b) The specimen elongation at force of 15 kNc) The reduction in specimen diameter at force of 15 kNF.TavgAHooke's lawd) The maximum Force where the specimen is brokene) The Force at Yield point.o = EeElongationALE =50070Tensile strength450 MPa (65,000 ры)60Poisson's ratio400dɛydexdetransV = -1050de arialpsiMPa 40300E = 2G (1+ v)4020030Yield strength30200250 MPa (36,00 psi)20100-2010This figure is thedetail figure from"Stress - Strain100100.005diagram" with strainrange from 0 - 0.0050.100.200.300.40Stress (MPa)Stress (10 psi

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3. Stress – Strain Behavior Consider the brass alloy for which the stress-strain behavior is shown in figure Formula below. A cylinder specimen of this material 10 mm in diameter and 120 mm long is Tensile stress: pulled in tension with force of 15 kN. The Poisson's ratio is 0.35. Compute : F. A Shear stress a) The Tension stress of 15 kN force. b) The specimen elongation at force of 15 kN c) The reduction in specimen diameter at force of 15 kN d) The maximum Force where the specimen is broken e) The Force at Yield point. F, Favg A Hooke's law o = Ee 500 Elongation 70 Tensile strength 450 MPa (65,000 ра) AL E = L 60 Poisson's ratio 400 dzy dex deuans 10° psi 50 de sial MPa 40 300 E = 2G (1+ v) 30 200 Yield strength 250 MPa (36.000 psi) 200 20 100 20 10 This figure is the detail figure from "Stress - Strain 100 10 0.005 diagram" with strain range from 0 - 0.005 0.10 0.20 0.30 0.40 Stress (MPa) (sd ot) saas

3. Stress – Strain Behavior Consider the brass alloy for which the stress-strain behavior is shown in figure Formula below. A cylinder specimen of this material 10 mm in diameter and 120 mm long is Tensile stress: pulled in tension with force of 15 kN. The Poisson's ratio is 0.35. Compute : F. A Shear stress a) The Tension stress of 15 kN force. b) The specimen elongation at force of 15 kN c) The reduction in specimen diameter at force of 15 kN d) The maximum Force where the specimen is broken e) The Force at Yield point. F, Favg A Hooke's law o = Ee 500 Elongation 70 Tensile strength 450 MPa (65,000 ра) AL E = L 60 Poisson's ratio 400 dzy dex deuans 10° psi 50 de sial MPa 40 300 E = 2G (1+ v) 30 200 Yield strength 250 MPa (36.000 psi) 200 20 100 20 10 This figure is the detail figure from "Stress - Strain 100 10 0.005 diagram" with strain range from 0 - 0.005 0.10 0.20 0.30 0.40 Stress (MPa) (sd ot) saas

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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